DESCRIPTION:

Compare two arrays and return a new array with any items only found in one of the two given arrays, but not both. In other words, return the symmetric difference of the two arrays.

Examples

[1, 2, 3, 5], [1, 2, 3, 4, 5]
// should return [4].

[1, "calf", 3, "piglet"], [1, "calf", 3, 4]
// should return ["piglet", 4].

My approach for solving this problem:

• concat the 2 arrayes.
• add every item into new obj
• if it's unquie get vlaue of 1 and get 2 if it's not.
• if not return sliced str with num length.
• filter item with value of 1 then map the number.

My solution:

function diffArray(arr1, arr2) {
    let newArr = arr1.concat(arr2)
    let numObj = {}
    newArr.forEach((item)=>{
      numObj[item] = numObj[item] ? numObj[item] +1:1
    })

    return Object.keys(numObj)
    .filter((item)=>{
      return numObj[item] == 1
    })
    .map((num)=>{
      if(!isNaN(num)){
          return +num
      }
      return num
    })
  }

  diffArray([1, "calf", 3, "piglet"], [1, "calf", 3, 4]);

freeCodeCamp approach for solving this problem:

• concat the 2 arrays.
• filter if item includes in arr1 or arr2 and return him.

freeCodeCamp solution:

function diffArray(arr1, arr2) {
  return arr1
  .concat(arr2)
  .filter(item => !arr1.includes(item) || !arr2.includes(item));
}

diffArray([1, "calf", 3, "piglet"], [1, "calf", 3, 4]);

Any tips or edit are most welcome. share it with me on the comments. Thanks for being here!

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